MA3H6 Exercise sheet 4. Please let me know if any of the problems are unclear or have typos. Please turn in a single worked exercise | write your name, the date, and the problem you are solving at the top of the page. If you collaborate with other students, please also include their names. Exercise 4.1. [Hatcher page 147, the splitting lemma.] Suppose that C∗ is a chain complex. If H∗(C) = 0 we call C∗ an exact sequence. (Equivalently, Zn = Bn for all n.) If, additionally, C∗ has at most three non-zero terms we call C∗ a short exact sequence. Show that every non-zero term is adjacent to another such. Now suppose p 0 ! A !i B ! C ! 0 is a short exact sequence of Abelian groups. Show the following are equivalent. • There is a homomorphism r : B ! A so that ri = IdA. • There is a homomorphism s: C ! B so that ps = IdC . • B ∼= A ⊕ C and there is an isomorphism from the given sequence to the sequence 0 ! A ! A ⊕ C ! C ! 0 with the obvious inclusion and projection maps. Such a sequence is called split; the maps r and s are called a retraction and a section, respectively. Exercise 4.2. [Hatcher page 148.] With notation as in Exercise 4.1, show if C is free then the sequence is split. Exercise 4.3. 1. For any d × n matrix M of integers the Smith normal form of M is a d × n matrix D = [di;j] with di;j 2 N, with di;j = 0 (if i 6= j), with di;i evenly dividing di+1;i+1, and having matrices U 2 GL(d; Z) and V 2 GL(n; Z) so that D = UMV . Learn how to compute Smith normal form in your favourite computer algebra system. 2. Find an algorithm that, given any chain complex C∗ = fCn;@ng of finitely generated free Abelian groups, computes the homology groups H∗(C). (Hints: Recall Bn = Im(@n+1) and Zn = Ker(@n). The inclusion of Zn into Cn, and @n, together give a short exact sequence 0 ! Zn ! Cn ! Bn−1 ! 0. Since Bn−1 is free, this sequence ∼ splits. Choose a splitting Cn = Zn ⊕ Bn−1 extending the inclusion of Zn into Cn. ∼ This induces an isomorphism Cok(@n+1) = Hn ⊕ Bn−1.) 3. Show that the following sequence of groups and homomorphisms C∗ is a chain complex. M N 0 −−−! Z3 −−−! Z5 −−−! Z −−−! 0 2014-01-27 1 MA3H6 Exercise sheet 4. Here the homomorphisms M and N are given by the matrices 2 −6 −26 −82 3 6 0 4 3 7 6 7 M = 6 1 0 7 7 and N = −2 −2 −4 −2 −4 : 6 7 4 0 2 5 5 2 10 30 4. Take M, N, and C∗ as in part (3) above. Compute the Smith normal forms of M and N. Use the algorithm to compute H∗(C). Exercise 4.4. We say an exact sequence is very short if it has at most two non-zero terms. i. Show the central map of a very short exact sequence is an isomorphism. ii. Show a long exact sequence of free Abelian groups may be written as a direct sum of very short exact sequences. Exercise 4.5. [Roberts.] For each exact sequence of Abelian groups below, say as much as possible about the group G and the homomorphism α. i. 0 ! Z2 ! G ! Z ! 0 ii. 0 ! Z ! G ! Z3 ! 0 α iii. 0 ! Z ! Z ⊕ Z ! Z ⊕ Z2 ! 0 α iv. 0 ! G ! Z ! Z ! Z2 ! 0 α v. 0 ! Z3 ! G ! Z2 ! Z ! Z ! 0 Here we write Zn for the group Z=nZ. Exercise 4.6. [Medium.] Suppose B and D are finitely generated free Abelian groups, A < B and C < D are subgroups, and B=A ∼= D=C. Show the chain complexes 0 ! A ! B ! 0 and 0 ! C ! D ! 0 are chain homotopy equivalent. (This is a generalization of Exercise 3.10. Smith normal form may be useful. See also problem 43(b) on page 159 of Hatcher.) Exercise 4.7. [Hard.] Show chain complexes C∗ and D∗ of finitely generated free Abelian groups are chain homotopy equivalent if and only if they have isomorphic homology ∼ groups: H∗(C) = H∗(D). (Hints are available at MathOverflow, question number 10974. Exercise 4.6 may be useful. See also problem 43(a) on page 159 of Hatcher.) Exercise 4.8. [Medium.] Find two topological spaces X and Y , with isomorphic homology groups, that are not homotopy equivalent. (Thus Exercise 4.7 does not generalize to topological spaces.) State any theorems from Hatcher that you use; you will need to read ahead a bit. 2014-01-27 2.